inverse laplace (s-10)/(s^2-2s-3)

⇥1) Let's factorize in

ℝ

s^{2} - 2 s - 3

⇥Given the quadratic form

a s^{2} + b s + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = -2

and

c = -3

⇥

Δ = {(-2)}^{2} - 4 \cdot 1 \cdot (-3)

⇥ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

{(-2)}^{2}

(-2) \cdot (-2)

4

⇥

Δ = 4 - 4 \cdot (-3)

⇥ⓘ Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive:

- 4 \cdot (-3)

12

⇥

Δ = 4 + 12

⇥ⓘ Start from

4

and move forward

12

on the number line:

4 + 12

16

⇥

Δ = 16

⇥As

Δ > 0

, the equation has two roots:

⇥The quadratic formula gives

s_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-2) - \sqrt{16}}{2 \cdot 1}

⇥

s_{1} = \frac{2 - \sqrt{16}}{2}

⇥ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{16}

4

;

4^{2}

16

⇥

s_{1} = \frac{2 - 4}{2}

⇥ⓘ The numerator and the denominator are computed before the division:

2 - 4

-2

⇥

s_{1} = - \frac{2}{2}

⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{2}{2}

-1

⇥

s_{1} = -1

⇥The quadratic formula gives

s_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-2) + \sqrt{16}}{2 \cdot 1}

⇥

s_{2} = \frac{2 + \sqrt{16}}{2}

⇥ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{16}

4

;

4^{2}

16

⇥

s_{2} = \frac{2 + 4}{2}

⇥ⓘ The numerator and the denominator are computed before the division:

2 + 4

6

⇥

s_{2} = \frac{6}{2}

⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{6}{2}

3

⇥

s_{2} = 3

⇥Factors found:

(s + 1) \cdot (s - 3)

⇥2) You can perform a partial fraction decomposition:

⇥

\frac{s - 10}{(s + 1) \cdot (s - 3)} = \frac{A}{s + 1} + \frac{B}{s - 3}

⇥For the simple factor:

s + 1

and its root:

-1

, compute A by multiplying by the denominator:

s^{2} - 2 s - 3

and evaluating at the root (the other terms cancel out):

⇥3) Let's solve in

ℝ

(- 1 - 3) \cdot A = - 1 - 10

⇥ⓘ Parentheses are computed first:

- 1 - 3

-4

■Start from

-1

and move back

10

more on the number line, away from zero:

- 1 - 10

-11

⇥

- 4 A = -11

⇥

4 A = 11

⇥ⓘ Both sides are divided by

4

⇥Solution obtained:

A = \frac{11}{4}

⇥For the simple factor:

s - 3

and its root:

3

, the unknown B can be easily computed by multiplying by the denominator:

s^{2} - 2 s - 3

and evaluating at the root because the other terms cancel out:

⇥4) Let's solve in

ℝ

(3 + 1) \cdot B = 3 - 10

⇥ⓘ Parentheses are computed first:

3 + 1

4

■Start from

3

and move back

10

on the number line, going below zero:

3 - 10

-7

⇥

4 B = -7

⇥ⓘ Both sides are divided by

4

⇥Solution obtained:

B = - \frac{7}{4}

⇥5) Finally, you can substitute the values of the unknowns into the initial partial fraction decomposition:

⇥

\frac{s - 10}{(s + 1) \cdot (s - 3)} = \frac{11}{4 \cdot (s + 1)} - \frac{7}{4 \cdot (s - 3)}

Find the causal inverse transform, for t ≥ 0. By linearity, invert each partial fraction

Before applying the formula, normalize the pole:

\frac{11}{4 \cdot (s + 1)} = \frac{\frac{11}{4}}{s + 1}

The inverse Laplace formula for

\frac{1}{{(s - a)}^{n}}

t^{n - 1} \cdot \frac{\exp (a t)}{(n - 1)!}

; it gives for

\frac{\frac{11}{4}}{s + 1}

\frac{\frac{11}{4} \cdot \exp (-1 \cdot t)}{0!}

Before applying the formula, normalize the pole:

- \frac{7}{4 \cdot (s - 3)} = - \frac{\frac{7}{4}}{s - 3}

The inverse Laplace formula for

\frac{1}{{(s - a)}^{n}}

t^{n - 1} \cdot \frac{\exp (a t)}{(n - 1)!}

; it gives for

- \frac{\frac{7}{4}}{s - 3}

\frac{- \frac{7}{4} \cdot \exp (3 t)}{0!}

⇥

\frac{\frac{11}{4} \cdot \exp (-1 \cdot t)}{0!} + \frac{- \frac{7}{4} \cdot \exp (3 t)}{0!}

⇥ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{11}{4} \cdot \exp (-1 \cdot t)

\frac{11 \exp (-1 \cdot t)}{4}

- \frac{7}{4} \cdot \exp (3 t)

- \frac{7 \exp (3 t)}{4}

■Multiplying by -1 gives the opposite:

-1 \cdot t

- t

■By convention, the factorial of 0 is 1:

0!

1

⇥

\frac{\frac{11 \exp (- t)}{4}}{1} - \frac{\frac{7 \exp (3 t)}{4}}{1}

ⓘ The factor is moved to the numerator or the denominator by changing the sign of its exponent, for a nonzero base:

\frac{11 \exp (- t)}{4}

\frac{11}{4 \exp (t)}

▶Inverse Laplace transform:

\frac{11}{4 \exp (t)} - \frac{7 \exp (3 t)}{4}